A converging lens has a focal length of 16.8 cm. For object distances of (i) 84.

Question

A converging lens has a focal length of 16.8 cm. For object distances of (i) 84.0 cm and (ii) 5.60 cm, consider the following.

(a) Determine the location of each image.

(b) Is image (i) real or virtual?

Is image (ii) real or virtual?

(c) Is image (i) upright or inverted?

Is image (ii) upright or inverted?

(d) What is the magnification of the image?

image (i) |distance|     cm image (i) location     ---Select--- in front of the lens or behind the lens image (ii) |distance|     cm image (ii) location     ---Select--- in front of the lens or behind the lens

Explanation / Answer

Converging lens

f = 16.8 cm

(a) Location of image

(i) Do = 84 cm

1/f = 1/Do + 1/Di

1/16.8 = 1/84 + 1/Di

Di = 84*16.8 / ( 84 - 16.8) = 21 cm

Distance of image is 21 cm

Image is behind the lens

(ii)

Do = 5.6 cm

1/f = 1/Do + 1/Di

1/16.8 = 1/5.6 + 1/Di

Di = 5.6*16.8 / ( 5.6 - 16.8) = - 8.4 cm

Distance of image is 8.4 cm

Image is behind the lens

(b)

Image (i) is Real

Image (ii) is virtual

(C) Image (i) is inverted

Image (ii) is upright

(d) magnification of image(i) = - Di/Do = - 21/84 = - 0.25

  magnification of image(ii) = - Di/Do = 8.4 /5.6 = 1.5

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